Given the list of pKa values, which acid will have the largest amount of conjugate base present in the solution at equilibrium?
Acid: Pka:
H20 15.7
NH3 38
HCL -8
NH4+ 9.2
H2O
NH3
NH4+
HCl <- My answer
I believe that the answer to this is HCl. I converted the pKa values to Ka, then wrote the ionizaton equation, and then wrote the Ka expression and found HCl. What are we thinking?
I don't know what you're thinking but I'm thinking HCl also.
HCl
Based on the pKa values provided, the acid with the largest amount of conjugate base present in the solution at equilibrium will be the one with the highest pKa value.
In this case, NH3 has the highest pKa value of 38. Therefore, NH3 will have the largest amount of conjugate base (NH2-) present in the solution at equilibrium.
So the correct answer is NH3, not HCl.
To determine which acid will have the largest amount of conjugate base present in the solution at equilibrium, we need to compare the pKa values. The lower the pKa value, the stronger the acid.
Using the information provided:
Acid: Pka:
H2O 15.7
NH3 38
HCl -8
NH4+ 9.2
H2O has the highest pKa value of 15.7, making it a weaker acid relative to the others.
NH3 has a pKa value of 38, indicating that it is the weakest acid among the choices.
HCl has a negative pKa value of -8, making it the strongest acid among the choices.
NH4+ has a pKa value of 9.2, indicating that it is weaker than HCl but stronger than H2O.
Therefore, the acid with the largest amount of conjugate base present at equilibrium will be the one with the lowest pKa value, which is HCl. Hence, your answer is correct.